Question

In a Harmonic progression $T_5=\frac{1}{12}$ and $T_8=\frac{1}{15}$. Find the value of $64\times T_1$.

Answer 1

Given : $T_5=\frac{1}{12}$ and $T_8=\frac{1}{15}$

In Harmonic Progression, $n^{th}$ term $T_n=\frac{1}{a+(n-1)d}$

$⟹$ $T_5=\frac{1}{a+(5-1)d}=\frac{1}{12}$ and $T_8=\frac{1}{a+(8-1)d}=\frac{1}{15}$ where $a$ is first term and $d$ is common difference.

$⟹$ $a+4d=12$ ......... $\left(1\right)$ and $a+7d=15$ .......... $\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ simultaneously, we get

$3d=3$ $⟹$ $d=1$

Substituting $d=1$ in equation $\left(1\right)$ we get, $a=8$

$\therefore$ $a=8$ and $d=1$ ............ $\left(3\right)$

$\therefore$ $T_1=\frac{1}{8}$.

$\Rightarrow 64\times T_1=8$